Question

Two masses 6 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released.



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Answer 1

Explanation:

Let ‘a’ be the acceleration of the masses.

  Then for block m1, T – m1g = m1a → (1)

for block m2, m2g – T = m2 a → (2)

(1) + (2) (m2 – m1) g = (m1 + m2) a

⇒ a =  (12 − 6)10/12 + 6

 a = 3.33m/s

substituting in (1)

T – 6× 10 = 6× 3.33

⇒ T = 79.98 N.

Answer 2

Answer:

So, the accelerations of the masses of 6kg and 12 kg is $a=3.33\frac{m}{sec^{2} }$.

Explanation:

Given:

$m_{6}$ = 6kg

$m_{12}$ = 12kg

Here,

Let the tension in the inextensible string between the two masses is denoted by T.

Let the accelerations of the masses of 6kg and 12 kg is denoted by a.

The force exerted by gravity on a mass of 6kg is denoted by $F_{g6}$.

The force exerted by gravity on a mass of 12kg is denoted by $F_{g12}$.

The mass of 6kg is denoted by $m_{6}$.

The mass of 12kg is denoted by $m_{12}$.

The acceleration due to gravity is denoted by g.

Now,

Write the motion equation for the mass of 6kg,

$T-F_{g6}=m_{6} a$                        

$T-m_{6}g=m_{6}a \\T-6\times10=6\\T-60=6 a$               .........(1)

Then,

Write the motion equation for the mass of 12kg,

$F_{g12}-T=m_{12} a$

$m_{12}g -T=m_{12}a$                        

$12\times10-T=12a\\120-T=12a$           ..........(2)

On solving the equation 1 and 2,

we get,

$a=3.33\frac{m}{sec^{2} }$

T = 80N

So, the accelerations of the masses of 6kg and 12 kg is $a=3.33\frac{m}{sec^{2} }$.